An archer shoots a 54.6 g arrow horizontally at a target 14.0 m away. The arrow is 18.0 inches long. The 54.6 g arrow is aimed directly at the center of the target, but it hits 60.0 cm lower. What was the initial speed of the arrow? (Neglect air resistance.)
To find the initial speed of the arrow, we can use the horizontal motion of the arrow and the vertical motion of the arrow separately.
First, let's analyze the horizontal motion. Since the arrow is shot horizontally, the initial horizontal velocity of the arrow is constant throughout its motion. The horizontal distance it travels is given as 14.0 m.
Next, let's look at the vertical motion of the arrow. We know that the arrow hits 60.0 cm lower than the center of the target. Since we neglect air resistance, the only force acting on the arrow vertically is gravity, which causes the arrow to accelerate downwards at 9.8 m/s².
Using these information, we can find the time it takes for the arrow to hit the target using the vertical motion equation:
Δy = v₀ × t + 0.5 × g × t²,
where Δy is the displacement vertically (-60.0 cm = -0.60 m), v₀ is the initial vertical velocity (0 m/s since it's shot horizontally), g is the acceleration due to gravity (-9.8 m/s²), and t is the time.
Plugging in the known values, we get:
-0.60 m = 0 × t + 0.5 × (-9.8 m/s²) × t².
Simplifying the equation, we have:
-0.60 m = -4.9 m/s² × t².
Next, we can use the horizontal motion to find the time it takes for the arrow to travel 14.0 m. Since the horizontal velocity is constant, we can use the equation:
v = Δx / t,
where v is the horizontal velocity, Δx is the horizontal distance (14.0 m), and t is the time.
Rearranging the equation to solve for time:
t = Δx / v.
We already know that the horizontal velocity is constant and the same as the initial speed of the arrow. Therefore, the time it takes to travel 14.0 m is:
t = 14.0 m / v.
Now we can substitute this value for t in the equation obtained from the vertical motion:
-0.60 m = -4.9 m/s² × (14.0 m / v)².
Simplifying the equation further, we have:
-0.60 m = -4.9 m²/s² × (14.0 m)² / v².
To solve for v², we can isolate it by rearranging the equation:
v² = (-4.9 m²/s² × (14.0 m)²) / (-0.60 m).
Calculating the values, we have:
v² = 1.435 × 10².
Taking the square root of both sides, we find:
v = √(1.435 × 10²) = 11.99 m/s.
Therefore, the initial speed of the arrow is approximately 11.99 m/s.